Science and engineering math: Difference equation, non-homogeneous

[chatterbug219]

Homework Statement

Solve the difference equation: a_n+2 - 5a_n+1 + 6a_n = 4ⁿ
Subject to a₀ = 0 & a₁ = 1

Homework Equations

a_n = Arⁿ

The Attempt at a Solution

I got two solutions for the first part for when it is homogeneous by substituting a_n = Arⁿ into the equation and solving for r.
a_n = A(3)ⁿ
a_n = A(2)ⁿ

I just don't know where to start for non-homogeneous. I tried a_n = B*4ⁿ and I got B*4ⁿ⁺² -5B*4ⁿ⁺¹ + 6B*4ⁿ = 4ⁿ but I don't know where to go from there.

 

[tiny-tim]

welcome to pf!

hi chatterbug219! welcome to pf!

I tried a_n = B*4ⁿ and I got B*4ⁿ⁺² -5B*4ⁿ⁺¹ + 6B*4ⁿ = 4ⁿ but I don't know where to go from there.

erm …

4ⁿ⁺²/4ⁿ = … ? ​

 

[chatterbug219]

So B = 1/2
And plugging that in...
a_n = (1/2)(4ⁿ) for the particular solution
Making the general solution:
a_n = A(2)ⁿ + (1/2)(4ⁿ)
Right?

 

[tiny-tim]

hi chatterbug219!

(just got up :zzz:)

… Making the general solution:
a_n = A(2)ⁿ + (1/2)(4ⁿ)
Right?

you mean a_n = A(2)ⁿ + B(3)ⁿ + (1/2)(4ⁿ)

(or you could write it 2ⁿ(2^n-1 + A) + 3ⁿB )

 

[chatterbug219]

No...if you plug in the initial conditions they final answers don't match up...
Only 2 works

 

[tiny-tim]

yes, but to solve for the initial condition, you also need to find A

 

[Ray Vickson]

Homework Statement

Solve the difference equation: a_n+2 - 5a_n+1 + 6a_n = 4ⁿ
Subject to a₀ = 0 & a₁ = 1

Homework Equations

a_n = Arⁿ

The Attempt at a Solution

I got two solutions for the first part for when it is homogeneous by substituting a_n = Arⁿ into the equation and solving for r.
a_n = A(3)ⁿ
a_n = A(2)ⁿ

I just don't know where to start for non-homogeneous. I tried a_n = B*4ⁿ and I got B*4ⁿ⁺² -5B*4ⁿ⁺¹ + 6B*4ⁿ = 4ⁿ but I don't know where to go from there.

The simplest way is to determine the generating function A(z) = \sum_{n=0}^{\infty} a_n z^n, then invert it to find \{ a_n \}. Google "difference equations" to see hundreds of examples.

RGV